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Let $f(x)= \begin{cases}x^2+k, & \text { when } x \geq 0 \\ -x^2-k, & \text { when } x \lt 0\end{cases}$. If the function $f(x)$ be continuous at $x=0$, then $k=$
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Verified Answer
The correct answer is:
$0$
We have,
$\begin{array}{ll}x^2+k & \text { when, } x \geq 0 \\
-x^2-k & \text { when, } x \leq 0\end{array}$
Given that,
L. H. L = R.H.L = f (x)
At point $x=0$
$\begin{aligned}& x^2+k=-x^2-k=0 \\
& \Rightarrow x^2+k=0 \\
& \Rightarrow k=-x^2 \\
& \Rightarrow k=0\end{aligned}$
$\begin{array}{ll}x^2+k & \text { when, } x \geq 0 \\
-x^2-k & \text { when, } x \leq 0\end{array}$
Given that,
L. H. L = R.H.L = f (x)
At point $x=0$
$\begin{aligned}& x^2+k=-x^2-k=0 \\
& \Rightarrow x^2+k=0 \\
& \Rightarrow k=-x^2 \\
& \Rightarrow k=0\end{aligned}$
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